Proof of the Fundamental Theorem of Space curves using Rigid Transformation by Peter Baxandall ( Vector Calculus ) I am reading Vector calculus by Peter Baxandall which proves the Fundamental theorem of Space curves ( Curves with equal torsion and curvature are identical except probably their position) in the following manner:


In the proof, the author says : Pick any $p \in E$. Hold $C_g$ fixed and move $C_h$ rigidly in $\Bbb R^3$ until $T_h(p) = T_g(p) , \cdots$. I do not see very clearly the motive and mechanism by which the author is able to do so. I understand rigid transformation as something which preserves the length of the curve. However, we may even have to employ rotation to make the unit tangent vector $T_g$ and $T_h$ the same. But, in the last line, he ultimately says that $C_h$ is a translation of $C_g$.


Also, I couldn't find where has the author used the fact that the torsion and curvatures of the two curves are equal.  $$\phi = T_g \cdot T_h + N_g \cdot N_h + B_g \cdot B_h \\ \implies \phi' = T_g' \cdot T_h + T_g \cdot T_h' + N_g' \cdot N_h + N_g \cdot N_h' +  B_g' \cdot B_h + B_g \cdot B_h'$$. But since, we have already have : $T_g=T_h,N_g=N_h,B_g=B_h$, thus : $T_g⋅T_h'=0=T_g'⋅T_h$. Similarily, for others, each dot product turns out to be $0$. We haven't seemed to use the fact that the torsions and curvatures of the two curves are equal ?

Could someone please explain what's actually going on. Thanks a lot!
NOTE : $T,N,B$ represent the tangent, normal and bi-normal unit - vector respectively
 A: The statement is that $C_g$ and $C_h$ are "equal, up to a movement". In his proof the author  replaces $C_h$ by a congruent copy (again denoted by $C_h$) in the following way: He chooses a $p\in E$ and applies a rotation $R$ of ${\mathbb R}^3$ such that the original orthonormal triple $\bigl(T_h(p),N_h(p),B_h(p)\bigr)$ is mapped to the triple $\bigl(T_g(p),N_g(p),B_g(p)\bigr)$. When this constant rotation $R$ is applied to $C_h$ the curve $R(C_h)=:C_h$ does not yet coincide with $C_g$, but is (in fact) a translate of $C_g$. When you want you can apply in addition a translation $A$ such that $(A\circ R)(h(p))=g(p)$ , but it is not necessary. As readers we accept without further ado that the moved curve $C_h$ is congruent to the original $C_h$.
The hard part of the proof then consists in showing that the new $C_h$ is congruent to $C_g$. Here the Frenet formulas are used. You should actually compute $\phi'$ in order to see that the equality of $s\mapsto\kappa(s)$ and $s\mapsto\tau(s)$ for the two curves plays a role in showing that $\phi'=0$:
$$\eqalign{\phi'&=(T_g\cdot T_h+N_g\cdot N_h+B_g\cdot B_h)'\cr
&=T_g'\cdot T_h+T_g\cdot T_h'+N_g'\cdot N_h+N_g\cdot N_h'+B_g'\cdot B_h+B_g\cdot B_h')\cr
&=\kappa N_g\cdot T_h+\kappa T_g\cdot N_h+(-\kappa T_g+\tau B_g)\cdot N_h+(-\kappa T_h+\tau B_h)\cdot N_g-\tau N_g\cdot B_h-\tau B_g\cdot N_h\cr &=0\ .\cr}$$
In the end the "equality" of $C_g$ and $C_h$ comes from the uniqueness part for the solution of ODEs.
